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3 years ago

Please help me with 5th question

Mathematics

2 answers:

aleksley [76]3 years ago

7 0

Answer:

3 kg is the answer

hope it helps you

Sidana [21]3 years ago

3 0

Answer:

uog cougccutxcouxi tdddpfd tdosiduo

Step-by-step explanation:

গ cuogchlicglici xiyxryfkcyricgilxiygciyrcuoxucy

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Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 24 students, she finds 2 w

irina1246 [14]

Answer:

A 95% confidence interval for the proportion of students who eat cauliflower on Jane's campus is [0.012, 0.270].

Step-by-step explanation:

We are given that Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 24 students, she finds 2 who eat cauliflower.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of students who eat cauliflower

n = sample of students

p = population proportion of students who eat cauliflower

Here for constructing a 95% confidence interval we have used a One-sample z-test for proportions.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

Now, in Agresti and Coull's method; the sample size and the sample proportion is calculated as;

$n = n + Z^{2}__(\frac{_\alpha}{2})$

n = $24 + 1.96^{2}$ = 27.842

$\hat p = \frac{x+\frac{Z^{2}__(\frac{\alpha}{2}_) }{2} }{n}$ = $\hat p = \frac{2+\frac{1.96^{2} }{2} }{27.842}$ = 0.141

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.141 -1.96 \times {\sqrt{\frac{0.141(1-0.141)}{27.842} } }$ , $0.141 +1.96 \times {\sqrt{\frac{0.141(1-0.141)}{27.842} } }$ ]

= [0.012, 0.270]

Therefore, a 95% confidence interval for the proportion of students who eat cauliflower on Jane's campus [0.012, 0.270].

The interpretation of the above confidence interval is that we are 95% confident that the proportion of students who eat cauliflower on Jane's campus is between 0.012 and 0.270.

7 0

2 years ago

If vx=11u and vz=u+30 find the value of u that makes the Quadrilateral WXYZ a parallelogram

stiv31 [10]

This answer is quadrant 3 because if you do the math then you end up in quadrant 3

6 0

2 years ago

Intersection point of Y=logx and y=1/2log(x+1)

GalinKa [24]

Answer:

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

The Problem:

What is the intersection point of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ ?

Step-by-step explanation:

To find the intersection of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ , we will need to find when they have a common point; when their $x$ and $y$ are the same.